If $z = 3+i$, find the values of n for which $Im(z^n)=0$

Question

What I got is: $\sin(0.322n) = 0 $

Hence: $0.322n = \pi $
$n = 9.76$

and $0.322n = 2\pi$
$n = 19.5$

and $0.322n = 3pi $
$n = 29.27$ Hence set of n values is: $\pm[9.76,19.5,29.27....]$

This is what the Oxford book that I'm using has given as a solution:

Answer 1

Let $y_n = \operatorname{Im}((3+i)^n)$. Then $2y_n = (3+i)^n - (3-i)^n$ and so $y_{n+2}=6y_{n+1}-10y_n$ because $3\pm i$ are roots of $t^2=6t-10$. The first few values of $y_n$ are $$ \begin{array}{c} n & 0 & 1 & 2 & 3 & 4 \\ y_n & 0 &1 & 6 & 26 & 96 \end{array} $$ It follows by induction from $y_{n+2}=6y_{n+1}-10y_n$ that $y_n \equiv 6 \bmod 10$ for $n\ge 2$. Thus, $y_n=0$ iff $n=0$.

So, the question is actually interesting, even if the solution given in the book is misguided and wrong.